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标题: 問prep2 98 [打印本页]

作者: songxuebest 时间: 2010-11-2 07:16 标题: 問prep2 98

98.
The violent crime rate (number of violent crimes per 1,000 residents) in Meadowbrook is 60 percent higher now than it was four years ago. The corresponding increase for Parkdale is only 10 percent. These figures support the conclusion that residents of Meadowbrook are more likely to become victims of violent crime than are residents of Parkdale.

The argument above is flawed because it fails to take into account

A. changes in the population density of both Parkdale and Meadowbrook over the past four years

B. how the rate of population growth in Meadowbrook over the past four years compares to the corresponding rate for Parkdale

C. the ratio of violent to nonviolent crimes committed during the past four years in Meadowbrook and Parkdale

D. the violent crime rates in Meadowbrook and Parkdale four years ago

D我明白爲什麽對，但是A爲什麽不對？
如果P增加了10%，M增加了60%而同時M的人口基數增大的量比P要大很多呢？

哎？看完好像有點明白了...人口密度和人口基數好像是兩回事。

如果這題A改成“人口數量”，我覺得A就也說得通了吧？

作者: nimingxia90 时间: 2010-11-2 21:28

想了一下。A改成人口数量也不对。因为题目本来就是自己跟自己比，然后说M比P容易受害。打个比方，比如A和P人数一样的，A以前10个人犯罪，现在16个人犯罪，是不是增长了60%啊。然而呢，P以前是80个人犯罪，但是现在呢88个人犯罪，是不是怎长了10%啊。虽然60%》10%，但是还是P比较危险吧

作者: 卮言浅夏 时间: 2010-11-3 06:39

D啊
你要看人口密度和人口数量没直接关系

作者: rudder86 时间: 2010-11-4 06:44

假设：
Meadowbrook 犯罪率：四年前为10%；现在为10%+60%=70%
Parkdale 犯罪率：四年前为80%；现在为80%+10%=90%

你说哪里的人更容易犯罪？

这个解释不错~

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